Reversibility Thermodynamics Professor Lee Carkner Lecture 14

PAL # 13 Entropy  Work needed to power isentropic compressor  Input is saturated vapor at 160 kPa  From table A-12, v = , h 1 = , s 1 =  Output is superheated vapor, P = 900 kPa, s 2 = s 1 =  From table A-13, h 2 =  Mass flow rate  m’ = V’/v = / =  Get work from W = m’  h  W = (0.27)( ) =

Ideal Gas Entropy  From the first law and the relationships for work and enthalpy we developed:  We need temperature relations for du, dh, d v and dP   ds = c v dT/T + R d v/v  ds = c p dT/T + R dP/P

Solving for s  We can integrate these equations to get the change in entropy for any ideal gas process s 2 -s 1 = ∫ c p dT/T + R ln (P 2 /P 1 )   Either assume c is constant with T or tabulate results

Constant Specific Heats  If we assume c is constant: s 2 -s 1 = c v,ave ln (T 2 /T 1 ) + R ln ( v 2 / v 1 ) s 2 -s 1 = c p,ave ln (T 2 /T 1 ) + R ln (P 2 /P 1 )    since we are using an average 

Variable Specific Heats   s o = ∫ c p (T) dT/T (from absolute zero to T)   s o 2 – s o 1 = ∫ c p (T) dT/T (from 1 to 2)   s 2 – s 1 = s o 2 – s o 1 – R u ln(P 2 /P 1 )  Where s o 2 and s o 1 are given in the ideal gas tables (A17-A26)

Isentropic Ideal Gas   Approximately true for low friction, low heat processes  c v,ave ln (T 2 /T 1 ) + R ln ( v 2 / v 1 ) = 0 ln (T 2 /T 1 ) = -R/c v ln ( v 2 / v 1 ) ln (T 2 /T 1 ) = ln ( v 1 / v 2 ) R/cv (T 2 /T 1 ) = ( v 1 / v 2 ) k-1 

Isentropic Relations  We can write the relationships in different ways all involving the ratio of specific heats k (T 2 /T 1 ) = (P 2 /P 1 ) (k-1)/k (P 2 /P 1 ) = ( v 1 / v 2 ) k  Or more compactly    P v k =constant  Note that:   R/c v = k-1

Isentropic, Variable c   Given that a process is isentropic, we know something about its final state   P r = exp(s o /R)  T/P r = v r  (P 2 /P 1 ) = (P r2 /P r1 ) ( v 2 / v 1 ) = ( v r2 / v r1 )

Isentropic Work  We can find the work done by reversible steady flow systems in terms of the fluid properties   But we know   -  w = v dP + dke + dpe  w = -∫ v dP –  ke –  pe  Note that  ke and  pe are often zero

Bernoulli   We can also write out ke and pe as functions of z and V (velocity) -w = v (P 2 -P 1 ) + (V 2 2 -V 2 1 )/2 + g(z 2 -z 1 )   Called Bernoulli’s equation   Low density gas produces more work

Isentropic Efficiencies   The more the process deviates from isentropic, the more effort required to produce the work   The ratio is called the isentropic or adiabatic efficiency

Turbine  For a turbine we look at the difference between the actual (a) outlet properties and those of a isotropic process that ends at the same pressure (s)  T = w a / w s ≈ (h 1 – h 2a ) / (h 1 – h 2s )  

Compressor   C = w s / w a ≈ (h 2s – h 1 ) / (h 2a – h 1 )   For a pump the liquids are incompressible so:  P = w s / w a ≈ v (P 2 -P 1 ) / (h 2a – h 1 )

Nozzles  For a nozzle we compare the actual ke at the exit with the ke of an isentropic process ending at the same pressure   N = V 2 2a / V 2 2s ≈ (h 1 – h 2a ) / (h 1 – h 2s )   Can be up to 95%

Entropy Balance  The change of entropy for a system during a process is the sum of three things  S in =  S out =  S gen =  We can write as:   S sys is simply the difference between the initial and final states of the system  Can look up each, or is zero for isentropic processes

Entropy Transfer  Entropy is transferred only by heat or mass flow  For heat transfer:  S =  For mass flow:  S =  n.b. there is no entropy transfer due to work

Generating Entropy   friction, turbulence, mixing, etc.   S sys = S gen +  (Q/T)  S gen =  S sys + Q surr /T surr

S gen for Control Volumes  The rate of entropy for an open system: dS CV /dt =  (Q’/T) +  m’ i s i –  m’ e s e + S’ gen   Special cases:  Steady flow (dS CV /dt = 0)  S’ gen = -  (Q’/T) -  m’ i s i +  m’ e s e  Steady flow single stream:  S’ gen =  Steady flow, single stream, adiabatic:  S’ gen =

Next Time  Read:  Homework: Ch 7, P: 107, 120, Ch 8, P: 22, 30